带有共享整数对象的Java等待通知

我试图在多个线程中使用共享整数对象按顺序打印从1到10的数字。当使用共享对象作为AtomicInteger时，程序工作正常，但当使用普通整数对象时，程序抛出异常，我不知道为什么会发生这种情况

原子整数程序

import java.util.concurrent.atomic.AtomicInteger;  

public class ThreadingProblem {

  public static void main(String[] args) {  
    AtomicInteger sharedInt = new AtomicInteger(0);  
    Thread t1 = new Thread(new ThreadingPrintingTask(sharedInt), "PrinterThread");  
    Thread t2 = new Thread(new ThreadingIncrementingTask(sharedInt), "IncrementerThread");  
    t1.start();  
    t2.start();  
  }  
}  

class ThreadingPrintingTask implements Runnable {    

  private AtomicInteger sharedObject;  

  public ThreadingPrintingTask(AtomicInteger sharedObject) {  
    this.sharedObject = sharedObject;  
  }  

  @Override  
  public void run() {  
    try {  
      synchronized (sharedObject) {  
        while (true) {  
          sharedObject.wait();  
          System.out.println("Shared object value is: " + sharedObject);  
          sharedObject.notify();  
        }  
      }  
    }  
    catch (InterruptedException e) {  
    }  

  }  
}  

class ThreadingIncrementingTask implements Runnable {  

  private AtomicInteger sharedObject;  

  public ThreadingIncrementingTask(AtomicInteger sharedObject) {  
    this.sharedObject = sharedObject;  
  }  

  @Override  
  public void run() {  
    synchronized (sharedObject) {  
      while (this.sharedObject.get() < 10) {  
        this.sharedObject.incrementAndGet();  
        this.sharedObject.notify();  
        try {  
          this.sharedObject.wait();  
        }  
        catch (InterruptedException e) {  
          // TODO Auto-generated catch block  
          e.printStackTrace();  
        }  
      }  
    }  
  }  
}  

具有普通整数对象的程序

public class ThreadingProblem {

  public static void main(String[] args) {
    Integer sharedInt = new Integer(0);
    Thread t1 = new Thread(new ThreadingPrintingTask(sharedInt), "PrinterThread");
    Thread t2 = new Thread(new ThreadingIncrementingTask(sharedInt), "IncrementerThread");
    t1.start();
    t2.start();
  }
}

class ThreadingPrintingTask implements Runnable {

  private Integer sharedObject;

  public ThreadingPrintingTask(Integer sharedObject) {
    this.sharedObject = sharedObject;
  }

  @Override
  public void run() {
    try {
      synchronized (sharedObject) {
        while (true) {
          sharedObject.wait();
          System.out.println("Shared object value is: " + sharedObject);
          sharedObject.notify();
        }
      }
    }
    catch (InterruptedException e) {
    }

  }
}

class ThreadingIncrementingTask implements Runnable {

  private Integer sharedObject;

  public ThreadingIncrementingTask(Integer sharedObject) {
    this.sharedObject = sharedObject;
  }

  @Override
  public void run() {
    synchronized (sharedObject) {
      while (this.sharedObject < 10) {
        this.sharedObject++;
        this.sharedObject.notify();
        try {
          this.sharedObject.wait();
        }
        catch (InterruptedException e) {
          // TODO Auto-generated catch block
          e.printStackTrace();
        }
      }
    }
  }
}

---

this.sharedObject++不会执行您认为它会执行的操作

由于
Integer
是不可变的，因此它不能更改现有的共享对象。相反，此操作所做的是将值取消装箱到一个
int
，增加值，然后将其装箱回另一个
Integer
实例

因此，您的代码（几乎*）相当于以下代码：

int temp = this.sharedObject.intValue();
temp = temp + 1;
this.sharedObject = new Integer(temp);

由于此时您的对象不再是同一实例，因此您的
synchronized
块将不会与
wait（）
/
notify（）
调用对齐

请注意，这与
AtomicInteger
的原子性无关，只是与
++
操作符如何处理
整数有关

*实际上，您可能会得到一个缓存实例，而不是
newinteger（）
，但它仍然是一个不同的实例，因为它表示不同的
int
值

Exception in thread "IncrementerThread" java.lang.IllegalMonitorStateException
    at java.lang.Object.notify(Native Method)
    at com.itiviti.apps.catalys.shared.mock.ThreadingIncrementingTask.run(ThreadingProblem.java:52)
    at java.lang.Thread.run(Unknown Source)

int temp = this.sharedObject.intValue();
temp = temp + 1;
this.sharedObject = new Integer(temp);